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I was attempting this integral: $I = \int \dfrac{\arcsin x}{(1-x^2)^{3/2}} dx$. But I am getting the wrong answer.

My approach:

Let $\arcsin x = t \implies dx = \sqrt{1-x^2} dt$. So, $I =\frac{t}{\cos^2 t} dt = t × \tan t - \int \tan t dt = \dfrac{\arcsin x × x}{\sqrt {1-x^2}} + \ln (\cos t) + K $.

But apparently, the correct answer includes a negative sign in the first term, that is: $I = - \dfrac{\arcsin x × x}{\sqrt{1-x^2}}+ \ln (\sqrt{1-x^2}) + K $.

So, I want to know where that minus sign comes from.

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HeWhoMustBeNamed
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  • I think something went very wrong during your LaTeX implementation... –  Mar 19 '18 at 15:19
  • You know.... you can preview your latex before posting. And you really should. –  Mar 19 '18 at 15:20
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    I made a small start in the title. With "copy-paste" you can try the rest? – imranfat Mar 19 '18 at 15:22
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    Checking the original post, there seems to be a huge mistake in the above editing: the integrand originally was $$\frac{\arcsin x}{(1-x^2)^{3/2}}$$ . As it is now it appears as $$\frac{\arcsin x}{(1-x^2)^{3/\color{red}4}}$$ – DonAntonio Mar 19 '18 at 15:32
  • Sorry for this poor latex implementation. I don't know latex, so I pasted the syntax from other questions and this mess resulted. As you can see I have rectified it now. – HeWhoMustBeNamed Mar 19 '18 at 15:35
  • @MrReality What you call "the correct answer" is for the function with the exponent $;3/2;$ in the denominator , as was originally in your post. As it is now the answer is the mess that appears in the answer below... – DonAntonio Mar 19 '18 at 15:38
  • @DonAntonio, corrected that too. So, can you tell me which answer is correct: mine or the one with the minus sign? – HeWhoMustBeNamed Mar 19 '18 at 15:45
  • @MrReality Apparently, no minus sign must be there... – DonAntonio Mar 19 '18 at 15:52
  • As it stands now I don't think this question is inapt or the formatting is bad and so doesn't deserve this many downvotes. So can people upvote it to get it to "0" at least? – HeWhoMustBeNamed Mar 19 '18 at 15:53
  • @MrReality You're correct. Sometimes people in this site get excited and a frenzy of downvotes begins. I already upvoted your question. +1 – DonAntonio Mar 19 '18 at 15:54

  • "You're correct" - @DonAntonio, so you mean there is no negative sign in the first term of the solution, right?

     – HeWhoMustBeNamed Mar 19 '18 at 15:56
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    @MrReality No, there is none...but you can check this simply differentiating without a sign and with the sign... – DonAntonio Mar 19 '18 at 15:57
  • @DonAntonio, Thanks then. – HeWhoMustBeNamed Mar 19 '18 at 15:58

1 Answers1

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Whilst waiting for your LaTeX improvement...

$$\int \frac{\arcsin(x)}{(\sqrt{1-x^2})^{3/2}}\ \text{d}x = $$

$$ = -\frac{\pi \left(1-x^2\right)^{3/4} \, _3F_2\left(\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};1-x^2\right)}{2 \sqrt{2} \Gamma \left(\frac{5}{4}\right) \Gamma \left(\frac{7}{4}\right)}-2 x \sqrt[4]{1-x^2} \, _2F_1\left(\frac{3}{4},1;\frac{5}{4};1-x^2\right) \sin ^{-1}(x)$$

Where $_aF_b$ are the hypergeometric functions, and $\Gamma(\cdot)$ is the Euler Gamma function.